Math / Data & Statistics

QuestionSave
A random sample of male college baseball players and a random sample of male college soccer players were obtained independently and weighed. The accompanying table shows the unstacked weights (in pounds). The distributions of both data sets suggest that the population distributions are roughly Normal. Determine whether the difference in means is significant, using a significance level of 0.05 .
Click the icon to view the data table.
Let $\mu_{\text {Baseball }}$ be the population mean weight (in pol male college soccer players. Determine the hypoth $\begin{array}{l} H_{0}: \mu_{\text {Baseball }}-\mu_{\text {Soccer }}=0 \\ H_{\mathrm{a}}: \mu_{\text {Baseball }}-\mu_{\text {Soccer }} \neq 0 \end{array}$
Find the test statistic for this test. $\mathrm{t}=\square$ (Round to two decimal places as needed.) $\square$
Get more help -
Weights (in pounds) \begin{tabular}{|cccc|} \hline Baseball & Soccer & Baseball & Full Data Set \\ 184 & 164 & 180 & 153 \\ 193 & 186 & 206 & 165 \\ 186 & 183 & 192 & 169 \\ 179 & 184 & 177 & 151 \\ 190 & 179 & 176 & 144 \\ 202 & 188 & 189 & 171 \\ 183 & 169 & 199 & 174 \\ 172 & 180 & 193 & 177 \\ 202 & 165 & 178 & 172 \\ 219 & 175 & 188 & 184 \\ 223 & 164 & 188 & 152 \\ 190 & 185 & 179 & 159 \\ 167 & 180 & & \\ \hline \end{tabular}
Print Done

Studdy Solution

STEP 1

What is this asking? We need to figure out if the average weights of college baseball and soccer players are really different, using some data, and a significance level of $0.05$ . Watch out! Don't mix up the data for the two groups!
Also, remember that a significance level of $0.05$ means we're willing to accept a $5\%$ chance of being wrong if we say the means are different.

STEP 2

1. Calculate the means and standard deviations.
2. Calculate the pooled standard deviation.
3. Calculate the t-statistic.

STEP 3

Let's calculate the mean weight for the baseball players.
Add up all the weights and divide by the number of players, which is 25. $\frac{184 + 193 + \dots + 179}{25} = \frac{4661}{25} = 186.44$ The mean weight for baseball players is $\mu_{Baseball} = 186.44$ pounds.
Now, let's calculate the standard deviation.
First, find the variance by calculating the average of the squared differences from the mean: $\frac{(184 - 186.44)^2 + (193 - 186.44)^2 + \dots + (179 - 186.44)^2}{25 - 1} = \frac{3299.36}{24} = 137.4733$ The standard deviation is the square root of the variance: $\sigma_{Baseball} = \sqrt{137.4733} \approx 11.72$

STEP 4

Let's calculate the mean weight for the soccer players.
Add up all the weights and divide by the number of players, which is 22. $\frac{164 + 186 + \dots + 159}{22} = \frac{3858}{22} = 175.36$ The mean weight for soccer players is $\mu_{Soccer} = 175.36$ pounds.
Now, let's calculate the standard deviation.
First, find the variance by calculating the average of the squared differences from the mean: $\frac{(164 - 175.36)^2 + (186 - 175.36)^2 + \dots + (159 - 175.36)^2}{22 - 1} = \frac{1621.909}{21} = 77.2338$ The standard deviation is the square root of the variance: $\sigma_{Soccer} = \sqrt{77.2338} \approx 8.79$

STEP 5

We need the pooled standard deviation to account for the different sample sizes.
The formula is: $s_p = \sqrt{\frac{(n_1 - 1) \cdot s_1^2 + (n_2 - 1) \cdot s_2^2}{n_1 + n_2 - 2}}$ Where $n_1$ and $n_2$ are the sample sizes, and $s_1$ and $s_2$ are the standard deviations.
Plugging in our values: $s_p = \sqrt{\frac{(25 - 1) \cdot (11.72)^2 + (22 - 1) \cdot (8.79)^2}{25 + 22 - 2}}$ $s_p = \sqrt{\frac{24 \cdot 137.3584 + 21 \cdot 77.2641}{45}}$ $s_p = \sqrt{\frac{3300.6016 + 1622.5461}{45}} = \sqrt{\frac{4923.1477}{45}} = \sqrt{109.4033} \approx 10.46$ So, our pooled standard deviation is $s_p \approx 10.46$ .

STEP 6

The t-statistic formula is: $t = \frac{(\mu_1 - \mu_2) - (\mu_{Baseball} - \mu_{Soccer})}{s_p \cdot \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}$ Since our null hypothesis is that the difference in means is zero ( $H_0: \mu_{Baseball} - \mu_{Soccer} = 0$ ), the formula simplifies to: $t = \frac{186.44 - 175.36}{10.46 \cdot \sqrt{\frac{1}{25} + \frac{1}{22}}}$ $t = \frac{11.08}{10.46 \cdot \sqrt{0.04 + 0.0455}} = \frac{11.08}{10.46 \cdot \sqrt{0.0855}} = \frac{11.08}{10.46 \cdot 0.2924} \approx \frac{11.08}{3.059} \approx 3.62$ Our t-statistic is approximately $3.62$ .

STEP 7

The t-statistic is $3.62$ .

Was this helpful?

Get Studdy

Scan QR code with your device to get Studdy on iOS or Android

QR code

Studdy solves anything!

Download the app

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Parents Influencer program Contact Policy Terms